$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}
end{align*}
$$

$$
begin{align*}
frac{V_p}{V_s}=frac{N_p}{N_s}
end{align*}
$$

As we have given that $N_s>N_p$, so must be $V_s>V_p$.
This means that the transformer increases the voltage.

$$
begin{align*}
frac{V_p}{V_s}=frac{N_p}{N_s}=frac{I_s}{I_p}
end{align*}
$$

As we see that the current of transformer secondary side is irreversibly proportional to a voltage, if the voltage is increased, the current is reduced.

$$
begin{align*}
frac{V_s}{V_p}=frac{N_s}{N_p}=frac{I_p}{I_s}
end{align*}
$$

So in our problem:

$$
bold{a)}
$$

$$
begin{align*}
N_p&=N\
N_s&=2cdot{N}\
frac{V_s}{V_p}&=frac{N_s}{N_p}\
frac{V_s}{V_p}&=frac{2cdot{N}}{N}
end{align*}
$$

$$
boxed{frac{V_s}{V_p}=2}
$$

$$
bold{b)}
$$

$$
begin{align*}
N_p&=N\
N_s&=2cdot{N}\
frac{I_s}{I_p}&=frac{N_p}{N_s}\
frac{I_s}{I_p}&=frac{N}{2cdot{N}}
end{align*}
$$

$$
boxed{frac{I_s}{I_p}=0.5}
$$

b) $frac{I_s}{I_p}=0.5$

$$
begin{align*}
I_{max}&=sqrt{2}cdot{I_{rms}}
end{align*}
$$

Let’s substitute and compute:

$$
begin{align*}
I_{max}&=sqrt{2}cdot{2.5text{ A}}
end{align*}
$$

$$
boxed{I_{max}=3.54text{ A}}
$$

$$
begin{align*}
V_{rms}&=frac{1}{sqrt{2}}cdot{V_{max}}
end{align*}
$$

Let’s substitute and compute:

$$
begin{align*}
V_{rms}&=frac{1}{sqrt{2}}cdot{5text{ V}}
end{align*}
$$

$$
boxed{V_{rms}=3.54text{ V}}
$$

$$
begin{align*}
P_{avg}&=I^2_{rms}cdot{R}
end{align*}
$$

Let’s substitute and compute:

$$
begin{align*}
P_{avg}&=(3.2text{ A})^2cdot{180text{ $Omega$}}
end{align*}
$$

$$
boxed{P_{avg}=1843.2text{ W}}
$$

$$
begin{align}
frac{V_s}{V_p}=frac{N_s}{N_p}
end{align}
$$

We have given:

$$
begin{align*}
N_p&=50\
N_s&=125\
V_p&=25text{ V}
end{align*}
$$

$bold{a)}$

As we have more loops at a secondary coil, the transformer increase the voltage so the voltage on a secondary coil is:
$$
boxed{text{Higher than $25text{ V}$.}}
$$

$bold{b)}$

Let’s express the voltage on the secondary coil and substitute:

$$
begin{align*}
V_s&=frac{N_s}{N_p}cdot{V_p}\
V_s&=frac{125}{50}cdot{5text{ V}}
end{align*}
$$

$$
boxed{V_s=62.5text{ V}}
$$

b) $V_s=62.5text{ V}$

$$
begin{align}
frac{I_s}{I_p}=frac{N_p}{N_s}
end{align}
$$

We have given:

$$
begin{align*}
N_p&=150\
N_s&=35\
I_p&=2.1text{ A}
end{align*}
$$

$bold{a)}$

As we have more loops at a primary coil, the transformer decrease the voltage, and increase the current at secondary side, so the current on a secondary coil is:
$$
boxed{text{Higher than $2.1text{ A}$.}}
$$

$bold{b)}$

Let’s express the current through the secondary coil and substitute:

$$
begin{align*}
I_s&=frac{N_p}{N_s}cdot{I_p}\
I_s&=frac{150}{35}cdot{2.1text{ A}}
end{align*}
$$

$$
boxed{I_s=9text{ A}}
$$

b) $I_s=9text{ A}$